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misc2 [2013/12/06 16:33] potthastmisc2 [2023/03/28 09:14] (current) – external edit 127.0.0.1
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 located at $x_2$ and $f_{2}$ is only influenced by $\varphi_1$ located at $x_1$. The matrix located at $x_2$ and $f_{2}$ is only influenced by $\varphi_1$ located at $x_1$. The matrix
 \begin{equation} \begin{equation}
 +\label{C example}
 C = \left( \begin{array}{cc} c_{11} & c_{12} \\ c_{21} & c_{22} \end{array} \right) C = \left( \begin{array}{cc} c_{11} & c_{12} \\ c_{21} & c_{22} \end{array} \right)
 := \left( \begin{array}{cc} 1 & 0 \\ 3 & 0 \end{array} \right) := \left( \begin{array}{cc} 1 & 0 \\ 3 & 0 \end{array} \right)
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 only.  only. 
  
-**Lemma.** An operator is //local// if and only if by reordering of the variables  +**Lemma.**  
-it can be transformed into a diagonal operator. +If we have a local operator for which each measusment is influenced by a different point 
 +$x_{j} \in \mathbb{R}^d$, then by reordering of the variables  
 +it can be transformed into a diagonal operator. In general, when a point influences two or more 
 +output variables, diagonalization by reordering is not possible.  
  
 //Remark.// A reordering operation is equivalent to the application of a permutation  //Remark.// A reordering operation is equivalent to the application of a permutation 
-matrix $P$. +matrix $P$, i.e. a matrix which has exactly one element 1 in each row and column, with  
 +all other elements zero
  
 //Proof.// We first assume that in the state space $X = \mathbb{R}^n$ each element belong //Proof.// We first assume that in the state space $X = \mathbb{R}^n$ each element belong
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 is influenced by the entries in the $\ell$-th row of the matrix $H$. Thus, this is local is influenced by the entries in the $\ell$-th row of the matrix $H$. Thus, this is local
 if and only if at most one of the entries is non-zero. This applies to every row, i.e. in  if and only if at most one of the entries is non-zero. This applies to every row, i.e. in 
-each row there is at most one non-zero element.  +each row there is at most one non-zero element. By the assumption that different measurements 
- +are influenced by different points, this means that there can be at most one nonzero entry 
-$\Box$ \\+in each column as well. But that means that the operator $H$ looks like a scaled version of  
 +a permutation matrix $P$, with scaling $0$ allowed. Clearly, by reordering we can make this 
 +into a diagonal matrix. In general, we take (\ref{C example}) as counter example,  
 +and the proof is complete $\Box$ \\
  
  
misc2.1386344018.txt.gz · Last modified: 2023/03/28 09:14 (external edit)